Let \(\Omega \) be a finite measurable space — finite and discrete, so that every subset is measurable — and let \(\mu \) be a finite measure on \(\Omega \), writing \(\mu ^{\mathbb {R}}\) for its real-valued mass function. Then for every subset \(S \subseteq \Omega \),

That is, the measure of \(S\) is the sum of the masses of its individual points.

Let \(\Omega \) be a finite discrete measurable space, \(\alpha \) a finite type, \(\mu \) a finite measure on \(\Omega \), \(Z : \Omega \to \alpha \) a function, and \(P\) a predicate on \(\alpha \). Writing \(\mu ^{\mathbb {R}}\) for the real-valued measure obtained from \(\mu \), the measure of the set of points whose image under \(Z\) satisfies \(P\) decomposes as a sum over the fibers of \(Z\):

Let \(\Omega \) be a finite measurable space, and let \(\mu \) and \(\nu \) be two measures on \(\Omega \). Then \(\mu \) is absolutely continuous with respect to \(\nu \), written \(\mu \ll \nu \), if and only if for every point \(\omega \in \Omega \) with \(\nu (\{ \omega \} ) = 0\) one also has \(\mu (\{ \omega \} ) = 0\).

Let \(\Omega \) be a finite measurable space, let \(\mu \) be a probability measure on \(\Omega \), and let \(f : \Omega \to \mathbb {R}\) be any real-valued function. Then

Let \(\Omega \) be a finite measurable space, \(\alpha \) a finite type on which every subset is measurable, \(\mu \) a finite measure on \(\Omega \), and \(Z : \Omega \to \alpha \) and \(f : \alpha \to \mathbb {R}\) any functions. Then

where \(\mu (Z^{-1}(\{ z\} ))\) denotes the real number giving the measure of the fiber \(Z^{-1}(\{ z\} ) = \{ \omega \in \Omega : Z(\omega ) = z\} \).

Let \(\Omega \) be a finite measurable space equipped with a finite measure \(\mu \), let \(\alpha \) be a finite set, let \(Z : \Omega \to \alpha \) be a measurable (\(\alpha \)-valued) map, and let \(S \subseteq \Omega \). Writing \(\mu ^{\mathbb {R}}\) for the real-valued measure and \(\mu [\, \cdot \mid A\, ]\) for the measure conditioned on an event \(A\), we have

so the total mass of \(S\) is recovered by summing, over the values \(z\) of \(Z\), the mass of the fiber \(Z^{-1}(\{ z\} )\) times the conditional mass of \(S\) given that fiber.

Let \(\Omega \) be a nonempty finite set, equipped with the uniform probability measure that assigns each point mass \(1/\left\lvert \Omega \right\rvert \), and let \(S \subseteq \Omega \). Then the measure of \(S\), as a real number, is

the number of elements of \(\Omega \) lying in \(S\) divided by the total number of elements of \(\Omega \).

Let \(\Omega \) be a nonempty finite set, equipped with its discrete measurable structure, and let \(S \subseteq \Omega \). Under the uniform probability measure on \(\Omega \), the measure of \(S\), regarded as a real number, equals

where \(\left\lvert S\right\rvert \) is the number of elements of \(\Omega \) lying in \(S\) and \(\left\lvert \Omega \right\rvert \) is the total number of elements of \(\Omega \).

If a type \(\Omega \) carries the structure of a finite probability space — that is, it is finite and discrete, and is equipped with a probability measure — then \(\Omega \) is nonempty.

Let \(\Omega \) be a finite probability space, write \(\mu \) for its canonical probability measure, and let \(p(\omega ) = \mu (\{ \omega \} )\) denote the associated probability mass function. Then for every subset \(S \subseteq \Omega \),

Let \(\Omega \) be a finite probability space, so that \(\Omega \) is finite and discrete and its canonical volume measure \(\mathbb {P}\) is a probability measure, and let \(\alpha \) be a finite type equipped with a bijection \(e : \Omega \xrightarrow {\sim } \alpha \). Then the family of singleton masses \(\bigl(\mathbb {P}(\{ e^{-1}(a)\} )\bigr)_{a \in \alpha }\) is summable with sum \(1\).

Let \(\Omega _1\) and \(\Omega _2\) be finite probability spaces, and give the product \(\Omega _1 \times \Omega _2\) its induced finite probability space structure. Then for every \(x \in \Omega _1\) and \(y \in \Omega _2\), the probability mass function of the product assigns to the pair \((x, y)\) the product of the individual masses; that is, its value at \((x, y)\) equals the value of the mass function of \(\Omega _1\) at \(x\) times the value of the mass function of \(\Omega _2\) at \(y\).

Let \(\Omega _1\) and \(\Omega _2\) be finite probability spaces, with respective probability measures \(\mathbb {P}_1\) and \(\mathbb {P}_2\), and let \(A \subseteq \Omega _1\) and \(B \subseteq \Omega _2\) be any subsets. Then, under the product probability measure \(\mathbb {P}\) on \(\Omega _1 \times \Omega _2\), the measure of the rectangle \(A \times B\) factors as

where each quantity is taken as a real number.

Let \(\Omega \) be a finite probability space and let \(\iota \) be a finite index set. If \((A_i)_{i \in \iota }\) is a pairwise disjoint family of subsets of \(\Omega \), then the probability of their union is the sum of their probabilities:

where \(\mathbb {P}(\, \cdot \, ) \in \mathbb {R}\) denotes the real-valued probability measure of \(\Omega \).

Let \(\Xi \) be a finite probability space and let \(\Omega \) be a measurable space in which every subset is measurable. For any map \(\varphi \colon \Xi \to \Omega \) and any finite set \(T \subseteq \Omega \), the probability that \(\varphi \) lands in \(T\) decomposes as a sum over the points of \(T\):

where \(\mathrm{volume}^{\mathbb {R}}\) denotes the real-valued measure of the underlying probability measure on \(\Xi \).

Let \(\Omega \) be a finite probability space, and let \(T\) be a finite subset of \(\Omega \). Writing \(\mathbb {P}\) for the real-valued probability measure on \(\Omega \), the measure of \(T\) is the sum of the measures of its singletons:

Let \(\Omega \) be a finite probability space, and let \(p(\omega )\) denote the probability mass it assigns to each point \(\omega \in \Omega \). Then for every function \(f \colon \Omega \to \mathbb {R}\),

so that the expectation of \(f\) under the canonical probability measure equals the finite \(p\)-weighted sum of its values.

Let \(\Omega \) be a finite probability space, and let \(f\colon \Omega \to \mathbb {R}\) be any real-valued function. Then the square of the expectation of \(f\) is at most the expectation of its square:

where both integrals are taken with respect to the underlying probability measure.

Let \(\iota \) be a finite index type and let \(\Omega \) be a finite probability space, so that the product \(\Omega ^{\iota }\) of copies of \(\Omega \) indexed by \(\iota \) carries the corresponding product probability measure. Fix an index \(i \in \iota \) and a function \(f : \Omega \to \mathbb {R}\). Then integrating the map \(\omega _{\bullet } \mapsto f(\omega _i)\) against the product measure on \(\Omega ^{\iota }\) gives the same value as integrating \(f\) against the measure on \(\Omega \):

Let \(\iota \) be a finite index set, and for each \(i \in \iota \) let \(\Omega _i\) be a finite probability space carrying its canonical probability measure. For any choice of subsets \(s_i \subseteq \Omega _i\), the product measure on the finite product \(\prod _{i \in \iota } \Omega _i\) assigns to the box \(\prod _{i \in \iota } s_i\) the value

where both sides are read as real numbers.

Let \(\Omega \) be a finite probability space and let \(S \subseteq \Omega \). Then the (real-valued) probability of \(S\) equals the integral over \(\Omega \) of the indicator function of \(S\), taken with value \(1\) on \(S\) and \(0\) elsewhere:

where \(\mathbf{1}_S \colon \Omega \to \mathbb {R}\) is the function equal to \(1\) at points of \(S\) and \(0\) at points outside \(S\).

Let \(\Omega \) be a finite probability space and let \(S \subseteq \Omega \) be any subset. Then the probability of \(S\) equals the integral of the indicator function of \(S\) against the probability measure:

where the left-hand side is the real number obtained from the measure of \(S\).

Let \(\Omega \) be a finite probability space, and for each \(\omega \in \Omega \) let \(p(\omega )\) denote the weight that the associated probability mass function assigns to \(\omega \), viewed as a real number. Then these weights sum to one:

Let \(\Omega \) be a finite probability space, and let \(\mu (\{ \omega \} )\) denote the probability mass that its associated probability mass function assigns to a point \(\omega \in \Omega \). Then this weight, taken as a real number, is nonnegative:

Let \(\Omega \) be a finite probability space, and let \(\mu \) denote its associated probability mass function, assigning to each point \(\omega \in \Omega \) the real weight \(\mu (\{ \omega \} ) \ge 0\). Then there exists a point \(\omega \in \Omega \) with \(\mu (\{ \omega \} ) \gt 0\).

Let \(\Omega \) be a finite probability space, and let \(f : \Omega \to \mathbb {R}\) be a function satisfying \(f(\omega ) \le c\) for every \(\omega \in \Omega \), where \(c \in \mathbb {R}\) is a constant. Then the integral of \(f\) with respect to the probability measure on \(\Omega \) satisfies \(\int _\Omega f \le c\).

Let \(\Omega \) be a finite probability space and let \(f : \Omega \to \mathbb {R}\) be a nonnegative function, so that \(f(\omega ) \ge 0\) for every \(\omega \in \Omega \). Then for every real \(\varepsilon \gt 0\), the probability that \(f\) attains at least \(\varepsilon \) is bounded by the mean of \(f\) divided by \(\varepsilon \):

Let \(\Omega \) be a finite probability space and let \(f : \Omega \to \mathbb {R}\) be a function. If there is a constant \(c \in \mathbb {R}\) with \(c \lt f(\omega )\) for every \(\omega \in \Omega \), then \(c \lt \int _\Omega f\), where the integral is taken with respect to the probability measure on \(\Omega \).